75x : 49 =7 28
2x - 5 =27
(x4)5 = x
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5 + ( x + 27 ) = 64
( x + 27 ) = 64 - 5 ( x + 27 ) = 59 x = 59 - 27 x = 32a) \(\Rightarrow x+27=59\Rightarrow x=32\)
b) \(\Rightarrow x-2=39\Rightarrow x=41\)
c) \(\Rightarrow x+5=-322\Rightarrow x=-327\)
d) \(\Rightarrow5x=35\Rightarrow x=7\)
e) \(\Rightarrow4\left(x-5\right)=56\Rightarrow x-5=14\Rightarrow x=19\)
f) \(\Rightarrow15+x=37\Rightarrow x=22\)
g) \(\Rightarrow7\left(13-x\right)=35\Rightarrow13-x=5\Rightarrow x=8\)
h) \(\Rightarrow10\left(x+1\right)=100\Rightarrow x+1=10\Rightarrow x=9\)
1) \(\left(3x+2\right)^2-4\\ =\left(3x+2\right)^2-2^2\\ =\left(3x+2-2\right)\left(3x+2+2\right)\\ =3x.\left(3x+4\right)\)
2) \(4x^2-25y^2=\left(2x\right)^2-\left(5y\right)^2=\left(2x-5y\right)\left(2x+5y\right)\)
3) \(4x^2-49=\left(2x\right)^2-7^2=\left(2x-7\right)\left(2x+7\right)\)
4) \(8z^3+27=\left(2z\right)^3+3^3=\left(2z+3\right)\left(4z^2+6z+9\right)\)
5) \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2\right)^2-\left(\dfrac{1}{2}\right)^2=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)
6) \(x^{32}-1\\ =\left(x^{16}\right)^2-1^2\\ =\left(x^{16}-1\right)\left(x^{16}+1\right)\\ =\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
1: \(\left(3x+2\right)^2-4=3x\left(3x+4\right)\)
2: \(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
3: \(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
4: \(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
5: \(\dfrac{9}{25}x^4-\dfrac{1}{4}=\left(\dfrac{3}{5}x^2-\dfrac{1}{2}\right)\left(\dfrac{3}{5}x^2+\dfrac{1}{2}\right)\)
a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)
=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)
=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)
b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)
=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)
=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)
c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)
=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)
=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)
=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)
=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)
=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)
75x : 49 = 728
=> 75x : 72 = 728
=> 75x - 2 = 28
=> 5x - 2 = 28
=> 5x = 30
=> x = 30 : 5
=> x = 6
b) 2x - 5 = 27
=> 2x = 27 + 5
=> 2x = 32
=> 2x = 25
=> x = 5
c) (x4)5 = x
=> x4.5 = x
=> x20 = x
=> x20 - x = 0
=> x(x19 - 1) = 0
=> \(\orbr{\begin{cases}x=0\\x^{19}-1=0\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^{19}=1\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x^{19}=1^{19}\end{cases}}\)=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
Vậy \(x\in\left\{-1;0;1\right\}\)
\(a,7^{5x}:49=7^{28}\)
\(7^{5x}:7^2=7^{28}\)
\(7^{5x-2}=7^{28}\)
\(\Rightarrow5x-2=28\)\(\Rightarrow5x=30\)\(\Rightarrow x=6\)
\(b,2^x-5=27\)\(\Rightarrow2^x=32\)\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(c,\left(x^4\right)^5=x\)
\(\Rightarrow x^{20}=x\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)